145. 二叉树的后序遍历 - Binary Tree Postorder Traversal
【后序遍历-递归】【后序遍历-迭代】【后序遍历-先序的变异】
给定一个二叉树，返回它的 后序 遍历。题目描述
Example:
输入: [1,null,2,3]
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输出: [3,2,1]
Solutions
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
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Solution 【后序遍历-递归】 ( 0ms)
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class Solution {
List<Integer> result =new ArrayList<Integer>();
public List<Integer> postorderTraversal(TreeNode root) {
loop(root);
return result;
}
/* 单独写递归循环，减少return的次数 */
public void loop(TreeNode node){
if(node==null)
return;
loop(node.left);
loop(node.right);
result.add(node.val);
}
}
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Solution 【后序遍历-迭代】 ( 2ms)
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class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if(root == null)
return result;
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode pre = null;
stack.push(root);
while(!stack.isEmpty()){
TreeNode curr = stack.peek();
if((curr.left == null && curr.right == null) ||
(pre != null && (pre == curr.left || pre == curr.right))){
result.add(curr.val);
pre = curr;
stack.pop();
}else{/* 先将右结点压栈, 再将左结点入栈 */
if(curr.right != null)
stack.push(curr.right);
if(curr.left != null)
stack.push(curr.left);
}
}
return result;
}
}
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Solution 【后序遍历-先序的变异】 ( 2ms)
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class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if(root == null)
return result;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.pop();
if(node.left != null) stack.push(node.left);//和传统先序遍历不一样，先将左结点入栈
if(node.right != null) stack.push(node.right);//后将右结点入栈
result.add(0,node.val); //逆序添加结点值
}
return result;
}
}
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Notes